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| TOPIC | Revision Videos | Mind Maps |
|---|---|---|
| General Organic Chemistry | Revise | Mindmap |
| Acidic and Basic Strength | Revise | Mindmap |
| Hydrocarbons | Revise | Mindmap |
| Optical Isomerism | Revise | Mindmap |
| Haloalkane & Haloarenes | Revise | Mindmap |
| Alcohol, Phenol and Ethers | Revise | Mindmap |
| Carbonyl Compounds | Revise | Mindmap |
| Aldol, Cannizzaro and Haloform Reaction | Revise | Mindmap |
| Name Reaction | Revise | Mindmap |
| Oxidation Reaction | Revise | Mindmap |
| Reduction Reaction | Revise | Mindmap |
| Carboxylic Acid | Revise | Mindmap |
| Nitrogen Containing Compounds | Revise | Mindmap |
| Bio-molecules | Revise | Mindmap |
| Polymers | Revise | Mindmap |
| Chemistry in Everyday Life | Revise | Mindmap |
| Atomic Structure | Revise | Mindmap |
| Chemical Bonding | Revise | Mindmap |
| Periodic Table | Revise | Mindmap |
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Variation of vapour pressure with Temperature showing boiling point of solution is higher than that of pure solvent.
Solution of acetone and chloroform shows negative deviations from Raoult’s law because of the formation of the $H$ -bonds between acetone and chloroform molecules.
Total number of moles after association
$=1-\alpha+\frac{\alpha}{2}=1-\frac{\alpha}{2}$
$\therefore \quad i=\frac{1-\frac{\alpha}{2}}{1}=.504$
or $\quad \alpha=(1-.504) \times 2=.496 \times 2=.992$
percentage of association $=99.2 \%$
(a) In this method, the osmotic pressure is balanced by the external pressure so that there is no strain on the membrane.
(b) The concentration of the solution does not change because the entry of the solvent into the solution is prevented by the external pressure.
(c) The time taken for the measurement of osmotic pressure is much less in this method as compared to the other methods.
Relationships between V.P. and mole fraction for ideal solution
Intrinsic semi-conductors. The extrinsic semi-conductors are formed when impurities of certain elements are added to the insulators.
This process known as doping makes available electrons or holes for conductivity and thus, leads to two types of semi-conduction. These are called n-type semi-conductors and p-type semi-conductors.
n-type semi-conductors. n-type semiconductors are formed when impurity atoms containing more valence electrons than the atoms of the parent in insulator are introduced in it. These are called electron rich impurities. For example, when traces of phosphorus (a group 15 element) is added to pure silicon (a group 14 element) on electron of each phosphorus atom (has five valence electrons) becomes free because it is not involved in any bonding with silicon atom (has four valence electrons). The unbonded electrons are free to carry the electric current. Thus, silicon containing phosphorus as the impurity is n-type semi-conductor.
(ii) p-type semi-conductors. These are formed when impurity atoms containing lesser number of valence electrons than the atoms of the parent insulator element added to it. These are called electron deficient impurities. For example, when traces of boron. (a group 13 element) are added to pure silicon (a group 14 element) only three valence electrons of silicon may be involved in the bonding with the three valence electrons of boron. Since silicon atom has four valence electrons, this may result in creating vacancies or holes. These holes are responsible for electrical conductivity and will result in p-type semi-conductors.
$=52.51 g \mathrm{mol}^{-1}$
Thus, packing efficiency of simple cubic lattice is 52.4%
(b) Packing Efficiency of Body Centred Cubic (BCC) Structure
We know that in a body centred cubic unit cell there are two spheres per unit cell. Let a be the side of the unit cell and r be the radius of sphere.
volume of sphere $=\frac{4}{3} \pi r^{3}$
Volume occupied by two spheres $=2 \times \frac{4}{3} \pi r^{3}=\frac{8}{3} \pi r^{3}$
Face diagonal $A C=\sqrt{A B^{2}+B C^{2}}=\sqrt{2 a^{2}}$
In right angled triangle ACD
Body diagonal $A D=\sqrt{A C^{2}+C D^{2}}=\sqrt{2 a^{2}+a^{2}}=\sqrt{3} a$
In a body centred cubic unit cell, the spheres at the corners are not touching each other but are in contact with the sphere at the centre. As a result the body diagonal AD is equal to the four times the radius of the sphere.
$\therefore A D=\sqrt{3} a=4 r$
$a=\frac{4}{\sqrt{3}} r$
Volume of unit cell $=a^{3}=\left(\frac{4}{\sqrt{3}} r\right)^{3}=\frac{64}{3 \sqrt{3}} r^{3}$
Percentage of space occupied by spheres
$=\frac{\text { Volume of spheres }}{\text { Volume of cube }} \times 100$
$=\frac{\frac{8}{3} \pi r^{3}}{\frac{64}{3 \sqrt{3}} r^{3}} \times 100=\frac{8}{3} \times \frac{22}{7} \times \frac{3 \sqrt{3}}{64} \times 100=68 \%$
Thus, packing efficiency of a bcc arrangement is $68 \% .$
(c) Face centred cubic
In a cubic close packing, the unit cell is face centred cube.
In a face centred cubic unit cell there are 4 spheres per unit cell.
Let r be the radius of sphere and a be the edge length of the cube.
Volume of sphere $=\frac{4}{3} \pi r^{3}$
Volume of four spheres $=4 \times \frac{4}{3} \pi r^{3}=\frac{16}{3} \pi r^{3}$
In a face centred cubic unit cell, the spheres at corners are in contact with sphere at the face centre but are not touching each other. Therefore, face diagonal $\mathrm{AC}$ is equal to four times the radius of sphere.
$A C=4 r$
But in right angled triangle ABC (figure)
$A C=\sqrt{A B^{2}+B C^{2}}=\sqrt{a^{2}+a^{2}}=\sqrt{2} a$
$\therefore \sqrt{2} a=4 r$
$a=\frac{4 r}{\sqrt{2}}$
Volume of the unit cell $=a^{3}=\left(\frac{4 r}{\sqrt{3}}\right)^{3}=\frac{64}{2 \sqrt{2}} r^{3}$
Percentage of space occupied by spheres
$=\frac{\text { Volume of spheres }}{\text { Volume of cube }} \times 100=\frac{\frac{16}{3} \pi^{3}}{\frac{64}{2 \sqrt{2}} r^{3}} \times 100$
$=\frac{16}{3} \times \frac{22}{7} \times \frac{2 \sqrt{2}}{64} \times 100=74 \%$
Thus, packing efficiency of ccp arrangement is 74%. Similarly, packing efficiency of hcp is also 74%.
(iii) Interstitial Defects. This type of defect is caused due to the presence of ions in the normally vacant interstitial sites in the crystal. The ions occupying the interstitial sites are called interstitials. The formation of interstitial defects in determined by the size of the interstitial ion.
(iv) The “holes’ occupied by electrons are called F-centres (or colour centres) and are responsible for the colour of the compound and many other interesting properties. For example, the excess sodium in $\mathrm{NaCl}$ makes the crystal appear yellow, excess potassium in $K C l$ makes it violet and excess lithium in LiClmakes it pink. Greater the number of F-centres, greater is the intensity of colour. Solids containing F-centres are paramagnetic because the electrons occupying the “holes’ are unpaired.
(b) Volume of one unit cell
$=\mathrm{a}^{3}$
$=\left(3.54 \times 10^{-8} \mathrm{cm}\right)^{3}$
No. of unit cells $1.00 \mathrm{cm}^{3}$
$=\frac{1.00}{\left(3.54 \times 10^{-8}\right)^{3}}$
$=2.26 \times 10^{22}$
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As a result, the double bond character of the bond in carboxylic acids is greatly reduced as compared to that in aldehydes and ketones. In other words, the carbonyl group in carboxylic acids is not a true carbonyl group as in aldehydes and ketones and hence does not give some of the characteristic reactions of the carbonyl group. For example, unlike aldehydes and ketones, carboxylic acids do not form oximes, hydrazones, semicarbazones etc.
Thus, in $p$ -fluorobenzoic acid, $+R$ -effect outweighs the $-I-$ effect but is p-chlorobenzoic acid, it is the I-effect which outweighs the $+$ R-effect. Consequently acid is a weaker acid than p-chlorobenzoic acid.
Phenylacetylene
Therefore, carboxylic acids fail to give the characteristic reactions of the carbonyl groups as given by aldehydes and ketones.
In the trimethylbenzoic acid there are three electron releasing $\mathrm{CH}_{3}$ groups present. They tend to increase the electron density on the oxygen atom of $O-H$ bond in carboxylic acid. As a result, the acidic strength is reduced considerably and it is not so easy to esterify $2,4,6$ -trimethyl benzoic acid.
Thus, carbon-oxygen bond length in formate ion is the same for both the bonds while these are different in formic acid.
This means that benzoate ion is more stable than phenate ion and this makes benzoic acid stronger acid than phenol although the contributing structures for phenate ion (five) are more than for benzoate ion (two).
Therefore, acetyl chloride is a better acetylating agent than acetic acids.
Describe the following giving a chemical equation for each
(i) Transesterification,
(ii) Hofmann bromamide reaction
(c) $\quad \mathrm{NH}_{2}$ attached to $\mathrm{NH}$ is more nucleophile than $\mathrm{NH}_{2}$ attached to $\mathrm{C}=$ O group, therefore, reaction occurs through $\mathrm{NH}_{2}$ attached to NH to give the corresponding semicarbozone, i.e.,
In ketone we have two alkyl groups whereas in aldehydes only one alkyl group is attached to carbonyl carbon. Since there is lesser number of alkyl groups in case of aldehydes so they undergo nucleophillic addition more readily than ketones.
(b) Steric effect : The presence of more alkyl groups in case of ketones hinders the attack of nucleophile on the carbonyl group and so the ketones are less reactive than aldehydes.
(ii) Aromatic acids contain benzene ring and their molecular masses are much higher than aliphatic acids of acetic acid group. Aromatic, acids also involve stronger inter molecular attractions, thus they are solids.
$2-(2-$ bromophenyl) ethanal. It can also be named as
$2-(\text { o-bromophenyl })$ ethanal.
(b) 5 -chloro-3-ethyl-2-pentanone.
(c) $3,5$ -dimethyl phenyl ethanoate
The acetate ion obtained from acetic acid gets detabilised due to electron releasing effect of methyl group and is not formed
easily. Consequently dissociation of $\mathrm{CH}_{3} \mathrm{COOH}$ takes place to lesser extent.
(b) Formation of ammonia derivatives (oximes, hydrazone, semi-carbazone etc.) proceeds via the attack of carbonyl carbon with proton to form the conjugate acid.
Therefore, presence of an acid is a must for preparing these
derivatives $(p H<7)$ However, in strongly acid medium, the proton attacks the unshared pair of electrons on nitrogen to form the species
$R N^{+} H_{3}$ which cannot attack the carbonyl carbon.
$\stackrel{+}{H}+: N H_{2} R \longrightarrow N H_{3} R$
In basic medium, there is no protonation of carbonyl group and
hence no reaction.
$>C=O \frac{\text { basic medium }}{\text { medium }} \rightarrow$ No protonation Therefore, preparation of ammonia derivatives requires slightly
acidic medium $(p H=3.5)$ and its careful control is essential.
Grignard reagents, on the other hand, being less nucleophilic
add only to the more reactive $\mathrm{CO}_{2}$ but not to the less reactive resonance stabilized magnesium salt of the carboxylic acid from which carbxylic acid can be generated by hydrolysis with mineral acids.
Here a conc. solution of $N a O H$ is used which is not a reducing agent.
(b) The compound $[\mathrm{C}]$ along with ethanol has been obtained from compound [B] upon acid hydrolysis. since $[\mathrm{B}]$ is a $\beta$ ketoester, its structure may be given as follows:
(c) since $[\mathrm{B}]$ has been formed by the condensation of two moles of $[\mathrm{A}]$ in the presence of sodium ethoxide, the ester $[\mathrm{A}]$ is ethyl propanoate and it participates in Claisen Condensation as follows:
(b) When ethyl acetate is treated with aqueous $N a O H$ Sodium ethanoate and ethanol is formed ethanol on treatment with $I_{2}$ and $N a O H$ forms yellow ppt of iodoform.
When methyl acetate is treated with aqueous NaOH sodium ethanoate and Methanol is formed. Methanol does not respond to iodoform reaction
(c) When benzoic acid is warmed with ethyl alcohol and a little concentrated $\mathrm{H}_{2} \mathrm{SO}_{4},$ a fragrant odour of ethyl benzoate is obtained.
Benzaldehyde does not respond to this test.
Dehydration of C gives but-1-ene i.e., a product with four carbon atoms, therefore the other product B is also with four carbon atoms and has a terminal COOH group.
Accordingly, A is $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}$
(butyl butanoate)
$\mathrm{B}$ is $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}$ and
$\mathrm{C}$ is $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}$
The presence of three methyl group (which are electron repelling) reduces the polarity of $>C=O$ group on one hand and offer a steric hindrance to nucleophilic attack of $C N^{-}$ at group. Therefore, trimethyl cyclohexanone does not give good yield i.e., it gives very poor yield.
(b) The study of semicarbazide (NH3 derivative of urea,
reveal that two $\mathrm{N}-\mathrm{H}$ bonds of $\mathrm{NH}_{2}$ group away from group in semicarbazide are comparatively weaker than two $\mathrm{N}-\mathrm{H}$ bond of the other $-N H_{2}$ group closer to $>C=O$ group. Thus it
produces $H^{+}$ easily for protonation of group of aldehyde or ketone.
The nature of reaction, esterification of carboxylic acid, is reversible as ester formed react with water and produces reactants and thus an equilibrium is set up between reactants and products. The removal of ester or water shifts the equilibrium to right according to Le-Chatelier’s principle and more ester is formed.
(ii) Carboxyl group (-COOH) is electron withdrawing i.e. deactivating the benzene ring and thus electron density becomes very less at ortho and para position in comparison to meta position. Electrophiles (+vely charged species) find it easier to attack at meta position as there is higher electron density thus $-C O O H$ group is meta directing.
As there is positive charge on ortho and para position, electron density is higher at meta position and hence electrophilic substitution takes place at meta position. (Similarly nitro group is meta directing)
(iii) Carboxylic group (-COOH) in acids is highly polar and carboxylic acid, generally, exists as dimers containing two hydrogen bonds each as shown below.
These hydrogen bonds in carboxylic acids are stronger than those in alcohols. It is due to following two factors
(a) The $O-H$ bond of the carboxylic acids are more strongly polarised due to adjacent electron attracting $>C=$ O groups.
(b) The oxygen atom of the group $C=O$ in carboxylic acid is more negative as compared to the oxygen atom of the alcohol.
Thus carboxylic acids posses higher boiling points than corresponding alcohols of similar molecular masses.
reagents:
(i) Heating with excess of $H B r$
(ii) Dilute $K M n O_{4}$ in the presence of alkali
(iii) A dilute solution of $N a_{2} C O_{3}$
(iv) Phenyl hydrazine
(v) Tollen’s reagent
(b) The product of carbonyl compounds with semicarbazide is known as semicarbazone.
(c) Hemiacetal is a compound which contains an ether as well as alcohol functional group. For example, methoxyethanol is a hemiacetal.
(d) Ketal is a cyclic compound obtained by reaction of acetone with ethylene glycol.
(e) $2,4$ -DNP derivative: see properties of aldehydes/ ketones in section 12.7 of text.
(f) Acetal : Acetal are the product of aldehyde and monohydric alcohol
(g) Aldol is a condensation product of acetaldehyde in the presence of dil. $N a O H$
(h) Oxime : Aldehyde and Ketones react with hydroxylamine to form oximes.
(i) Schiff’s base : Schiff’s reagent is an aqueous solution of magenta or pink coloured rosaniline hydrochloride which has been decolourised by passing $\mathrm{SO}_{2}$ when aldehyde are treated with decolourised solution of schiff’s reagent, its pink or magenta colour is restored. This reaction is used as a test for aldehyde because ketones do not restore the pink colour of schiff reagent.
